What is the slope of the line tangent to $f(x) = -2x^{2}-x+6$ at $x = 1$ ?
The slope of the tangent line is $ \lim_{\Delta x \to 0} \frac{f(x + \Delta x) - f(x)}{\Delta x}$ $ = \lim_{\Delta x \to 0} \frac{(-2(x+\Delta x)^{2}-(x+\Delta x)+6) - (-2x^{2}-x+6)}{\Delta x}$ $ = \lim_{\Delta x \to 0} \frac{(-2(x^{2}+2x \Delta x+\Delta x^{2})-(x+\Delta x)+6) - (-2x^{2}-x+6)}{\Delta x}$ $ = \lim_{\Delta x \to 0} \frac{-2x^{2}-4(x \Delta x)-2\Delta x^{2}-x-\Delta x+6+2x^{2}+x-6}{\Delta x}$ $ = \lim_{\Delta x \to 0} \frac{-4(x \Delta x)-2\Delta x^{2}-\Delta x}{\Delta x}$ $ = \lim_{\Delta x \to 0} -4x-2(\Delta x)-1$ $ = -4x-1$ $ = (-4)(1)-1$ $ = -5$